KJSEA 2025 Maths Q26 — Simultaneous Equations (Cost of Books and Pens)

KJSEA 2025 Grade 9 Algebra

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The Question

“Regina bought four books and three pens for 315 shillings. Hermaize bought eight books and two pens for 530 shillings. Determine the cost of one book and one pen.”

1

Name the unknowns

Give each unknown quantity a letter so the word problem becomes algebra. Let B stand for the cost of one book and P stand for the cost of one pen, both in shillings. Everything that follows is built from these two symbols.

B=cost of one book,P=cost of one penB = \text{cost of one book}, \quad P = \text{cost of one pen}
2

Write an equation for each purchase

Turn each sentence into an equation. Regina's four books and three pens cost 315 shillings, and Hermaize's eight books and two pens cost 530 shillings. Each purchase gives you one equation linking B and P.

4B+3P=315(equation 1)4B + 3P = 315 \quad \text{(equation 1)}
8B+2P=530(equation 2)8B + 2P = 530 \quad \text{(equation 2)}
3

Match the book terms

To use elimination you want the number of books to be the same in both equations. Every term in equation 2 is even, so divide the whole equation by 2. This scales it down without changing what it means and makes the book term match equation 1.

8B+2P2=5302\frac{8B + 2P}{2} = \frac{530}{2}
4B+P=265(equation 3)4B + P = 265 \quad \text{(equation 3)}
4

Eliminate the books

Both equation 1 and equation 3 now contain four books, so subtracting one from the other cancels the books completely and leaves a single equation in P. Subtract equation 3 from equation 1.

(4B+3P)(4B+P)=315265(4B + 3P) - (4B + P) = 315 - 265
2P=50P=252P = 50 \Rightarrow P = 25

The book terms cancel because four books minus four books is zero, leaving only pens.

5

Find the cost of a book

Now that you know a pen costs 25 shillings, substitute that value back into equation 3 and solve for B. This is called back-substitution: use the answer you already have to unlock the one you still need.

4B+25=2654B + 25 = 265
4B=240B=604B = 240 \Rightarrow B = 60

Final Result

One book costs 60 shillings and one pen costs 25 shillings.

Why this method works

Elimination works because two straight-line equations that share a solution can be combined without losing that solution. Scaling equation 2 down by 2 keeps every point on its line unchanged, so it still holds the true prices. Once the book coefficients are identical, subtracting the equations removes B entirely, collapsing a two-unknown problem into a one-unknown problem you can solve directly. Substituting that value back recovers the second unknown, and because both original equations are still satisfied, the pair of prices is the unique answer that fits both purchases.

Check both purchases: four books and three pens give 4(60) + 3(25) = 315, and eight books and two pens give 8(60) + 2(25) = 530. Both match.